Q.Cyclopropane $\rightarrow$ Propene; this reaction follows first order kinetics. The rate constant at a particular temperature is $2.303\times10^{-2}\ \text{hour}^{-1}$. The initial concentration of cyclopropane is 0.25 M. What will be the concentration of cyclopropane after 1806 minutes? ($\log 2=0.3010$)

Step 1. Convert 1806 minutes to hours to match k's units: $t=1806/60=30.1$ hours.

Step 2. Using $k=\dfrac{2.303}{t}\log\dfrac{[A]_0}{[A]}$: $2.303\times10^{-2}=\dfrac{2.303}{30.1}\log\dfrac{[A]_0}{[A]}\Rightarrow\log\dfrac{[A]_0}{[A]}=\dfrac{2.303\times10^{-2}\times30.1}{2.303}=10^{-2}\times30.1=0.301$. …

Substitute the given k and t (converted to consistent hour units) into the first order rate law, recognise the resulting log v …

• Forgetting to convert 1806 minutes into hours before using an hour$^{-1}$ rate constant, which would m …